Can we construct free semi-lattice from free dist. lattice?
Dear All, I have a problem which I had thought very specialised, but actually can be stated very generally: - I have a category C with finite limits, and so I also have a category DLat(C) of distributive lattices which, lets say, has coequalizers. If free distibrutive lattices can be constructed (i.e. if there exists F:C->DLat(C) left adjoint to the forgetful functor) then do free semilattices exist? Thanks, Christopher Townsend (OU)
On Tue, 27 May 2003, Christopher Townsend wrote:
Dear All, I have a problem which I had thought very specialised, but actually can be stated very generally: -
I have a category C with finite limits, and so I also have a category DLat(C) of distributive lattices which, lets say, has coequalizers. If free distibrutive lattices can be constructed (i.e. if there exists F:C->DLat(C) left adjoint to the forgetful functor) then do free semilattices exist?
I suspect the answer is no, for irritatingly trivial reasons. If C is a pointed category (i.e. has a zero object), then any internal distributive lattice in C has its top and bottom elements equal, and so is degenerate (i.e. isomorphic to the terminal object 1). Hence the free-distributive-lattice functor exists, and is the constant functor with value 1. But I'm sure there must be examples of pointed, finitely complete categories which don't have a free-semilattice functor (though I have to admit I don't have one at my fingertips). Peter Johnstone
participants (2)
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Christopher Townsend -
Prof. Peter Johnstone